Đáp án:
\(\dfrac{{125}}{{27}}\)
Giải thích các bước giải:
\(\begin{array}{l}
P = 27{x^3} - 3.9{x^2}.y + 3.3x.{y^2} - {y^3} - 2{x^2} + {y^2} + {x^3} - 3.{x^2}.3y + 3x.9{y^2} - 27{y^3}\\
= 28{x^3} - 28{y^3} - 2{x^2} + {y^2} - 36{x^2}y + 36x{y^2}\\
Thay:x = \dfrac{1}{3};y = - \dfrac{1}{3}\\
\to P = 28.{\left( {\dfrac{1}{3}} \right)^3} - 28.{\left( { - \dfrac{1}{3}} \right)^3} - 2{\left( {\dfrac{1}{3}} \right)^2} + {\left( { - \dfrac{1}{3}} \right)^2} - 36.{\left( {\dfrac{1}{3}} \right)^2}.\left( { - \dfrac{1}{3}} \right) + 36.\left( {\dfrac{1}{3}} \right).{\left( { - \dfrac{1}{3}} \right)^2}\\
= \dfrac{{125}}{{27}}
\end{array}\)
